删除所有重复的出现，并在列表中保留一个:

test = [1, 1, 2, 3]

newlist = list(set(test))

print newlist

[1, 2, 3]

下面是我在Project Euler中使用的函数:

def removeOccurrences(e):
  return list(set(e))

有什么问题:

Motor=['1','2','2']
for i in Motor:
   if i != '2':
       print(i)
print(motor)

Numpy方法和对包含1.000.000个元素的列表/数组的计时:

计时:

In [10]: a.shape
Out[10]: (1000000,)

In [13]: len(lst)
Out[13]: 1000000

In [18]: %timeit a[a != 2]
100 loops, best of 3: 2.94 ms per loop

In [19]: %timeit [x for x in lst if x != 2]
10 loops, best of 3: 79.7 ms per loop

结论:numpy(在我的笔记本上)比列表理解方法快27倍

PS如果你想将常规的Python列表lst转换为numpy数组:

arr = np.array(lst)

设置:

import numpy as np
a = np.random.randint(0, 1000, 10**6)

In [10]: a.shape
Out[10]: (1000000,)

In [12]: lst = a.tolist()

In [13]: len(lst)
Out[13]: 1000000

检查:

In [14]: a[a != 2].shape
Out[14]: (998949,)

In [15]: len([x for x in lst if x != 2])
Out[15]: 998949

您可以将列表转换为numpy。数组，然后使用np.delete并传递该元素及其所有出现的元素的索引。

import numpy as np

my_list = [1, 2, 3, 4, 5, 6, 7, 3, 4, 5, 6, 7]
element_to_remove = 3
my_array = np.array(my_list)

indices = np.where(my_array == element_to_remove)
my_array = np.delete(my_array, indices)   
my_list = my_array.tolist()

print(my_list)

#output
[1, 2, 4, 5, 6, 7, 4, 5, 6, 7]

你可以使用列表推导式:

def remove_values_from_list(the_list, val):
   return [value for value in the_list if value != val]

x = [1, 2, 3, 4, 2, 2, 3]
x = remove_values_from_list(x, 2)
print x
# [1, 3, 4, 3]

更好的解决方案与列表理解

x = [ i for i in x if i!=2 ]